Election in Jacksonville, Florida, US
2019 Jacksonville mayoral election|
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Turnout | 24%[1] |
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The 2019 Jacksonville mayoral election was held on March 19, 2019, to elect the mayor of Jacksonville. Incumbent mayor Lenny Curry won a majority of votes to win a second term in office.[2]
Candidates
Republican Party
Declared
Democratic Party
While Democratic candidates did declare their candidacy, no Democratic candidates qualified for the mayoral election in 2019.
Declared
- Doreszell Cohen, founder of Citizens for Criminal Justice Reform[6]
- Yolanda Thornton, small business owner[7]
Declined
- Alvin Brown, former mayor of Jacksonville[8]
- Garrett Dennis, member of Jacksonville City Council[5][9]
Independents
Declared
- Omega Allen
- Connell Crooms, nonprofit director[10]
- Vishaun Grissett, independent consultant[11]